异步等待持久警告

Question

我得到警告“这个异步方法缺少'await'运算符，将同步运行。考虑使用'await'运算符来等待非阻塞API调用，或者使用'await task.run(...)‘来在后台线程上执行CPU绑定的工作。虽然这听起来很明显，但我认为我已经被覆盖了，因为我运行所有进程，并在我的processFiles方法中有一个await任务。Whenall（）。我需要用不同的方式来做这件事吗？如果有任何想法，我将不胜感激。

    using System;
    using System.Collections.Generic;
    using System.ComponentModel;
    using System.Data;
    using System.Data.SqlClient;
    using System.Drawing;
    using System.Linq;
    using System.Text;
    using System.Threading.Tasks;
    using System.Windows.Forms;
    using System.IO;


    namespace ProcessFiles
    {
        public partial class ProcessFiles : Form 
        {
            public ProcessFiles(int userLevel, int userID)
            {
            }

            private void btnProcess_Click(object sender, EventArgs e)
            {
                processFiles();

            }

            async Task<int> processFiles()
            {
                var processingTasks = new List<Task>();
                foreach (string fileName in listBox1.Items)
                {
                    processingTasks.Add(process012(fileName));
                    processingTasks.Add(process123(fileName));
                    processingTasks.Add(process234(fileName));

                }
                await Task.WhenAll(processingTasks);
                return (1);
            }


            async Task<int> process173(string fileName)
            {
                return (retVal);

            }

            async Task<int> process032(string fileName)
            {
                return (retVal);
            }

            async Task<int> process018(string fileName)
            {
                return (retVal);
            }

Answer 1

无需在processXXX方法中使用异步任务，只需返回Task 即可。您将在main processFiles（）方法中等待任务：

        private async void btnProcess_Click(object sender, EventArgs e)
        {
            await processFiles();
        }

        async Task<int> processFiles()
        {
            var processingTasks = new List<Task>();
            foreach (string fileName in listBox1.Items)
            {
                processingTasks.Add(process012(fileName));
                processingTasks.Add(process123(fileName));
                processingTasks.Add(process234(fileName));

            }
            await Task.WhenAll(processingTasks);
            return (1);
        }


        Task<int> process173(string fileName)
        {
            return Task.FromResult(retVal)
        }

        Task<int> process032(string fileName)
        {
            return Task.FromResult(retVal)
        }

        Task<int> process018(string fileName)
        {
           return Task.FromResult(retVal)
        }